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Calculating Power Transfer With Varying Impedences

Discussion in 'Mathematics and Physics' started by muso52, May 29, 2014.

  1. muso52

    muso52 Member

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    Location:
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    Hi People,

    If an amplifier delivers 3 watts into a load of matching impedence at 4ohms then 3 watts would be dissipated across the load . However what would be the result of doubling the impedence of the load to 8ohm? The power across the load would not be halved becauseI believe it is a logarithmic relationship. At a guess I think it would be something like 2.4watts but can any one tell me (show me) the equation please?

    Thanks in advance,

    Willy.
     
  2. N.Carsane

    N.Carsane New Member

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    (P=I*V AND I=V/R) P= (V^2)/R , i would assume the voltage output of the amp is the same so P is inversely linked to R therefore if if you double the Resistance, the power would halve. So the power would be be 1.5 Watts. That's my take on it...

    But... i'm not sure until we have readings, what is the max voltage output from the amp? And/or the current?
     
  3. Ratchit

    Ratchit Well-Known Member

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    Sure, I can. We have a total of 6 watts of power and 8 ohms of resistance. That makes an amplifier voltage of 4*sqrt(3) . By using the voltage divider method, we determine the voltage across the load is (8/12)*4*sqrt(3) . So the power of the voltage across the 8 ohm load is 8/3 watts.

    Ratch
     

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